$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$AD = AD$ [Common]
$BD = CD$ [Given]
$\therefore \Delta ABD \cong \Delta ACD$ [$SSS$ congruence criterion]
$(ii)$ In $\Delta ABP$ and $\Delta ACP$,we have:
$AB = AC$ [Given]
$\angle BAP = \angle CAP$ [Since $\Delta ABD \cong \Delta ACD$,their corresponding parts are equal]
$AP = AP$ [Common]
$\therefore \Delta ABP \cong \Delta ACP$ [$SAS$ congruence criterion]
$(iii)$ Since $\Delta ABP \cong \Delta ACP$,their corresponding parts are congruent.
$\Rightarrow \angle BAP = \angle CAP$,so $AP$ bisects $\angle A$.
Also,in $\Delta BDP$ and $\Delta CDP$:
$BD = CD$ [Given]
$DP = DP$ [Common]
$BP = CP$ [Since $\Delta ABP \cong \Delta ACP$,$BP = CP$ by $CPCT$]
$\therefore \Delta BDP \cong \Delta CDP$ [$SSS$ congruence criterion]
$\Rightarrow \angle BDP = \angle CDP$,so $AP$ bisects $\angle D$.
$(iv)$ Since $\Delta ABP \cong \Delta ACP$,we have $\angle APB = \angle APC$ [by $CPCT$].
Since $\angle APB + \angle APC = 180^\circ$ [Linear pair],
$\angle APB = \angle APC = 90^\circ$.
Also $BP = CP$ [by $CPCT$].
Thus,$AP$ is the perpendicular bisector of $BC$.